A circle with radius $6$ has a sector with a $\dfrac{9}{10}\pi$ radian central angle. What is the area of the sector? ${36\pi}$ $\color{#9D38BD}{\dfrac{9}{10}\pi}$ ${\dfrac{81}{5}\pi}$ ${6}$
Solution: First, calculate the area of the whole circle. Then the area of the sector is some fraction of the whole circle's area. $A_c = \pi r^2$ $A_c = \pi (6)^2$ $A_c = 36\pi$ The ratio between the sector's central angle $\theta$ and $2 \pi$ radians is equal to the ratio between the sector's area, $A_s$ , and the whole circle's area, $A_c$ $\dfrac{\theta}{2 \pi} = \dfrac{A_s}{A_c}$ $\dfrac{9}{10}\pi \div 2 \pi = \dfrac{A_s}{36\pi}$ $\dfrac{9}{20} = \dfrac{A_s}{36\pi}$ $\dfrac{9}{20} \times 36\pi = A_s$ $\dfrac{81}{5}\pi = A_s$